My apologies for the relatively long question, but I am trying to understand a step in a proof, which needs some preliminary explanation.
Let $K$ be a field and $I$ a prime ideal of $K[X_1,\ldots,X_n]$. I am trying to understand a proof of a theorem in which it is stated that $L$, the fraction field of $K[X_1,\ldots,X_n]/I$, is a finite algebraic extension of a purely transcedental extension of $K$.
Let $m$ be the maximal number of indices such that $K[X_{i_1},\ldots,X_{i_m}]\cap I = \emptyset$ and assume without loss of generality that those indices are exactly the first $m$. Then $K[X_1,\ldots,X_m]$ is a subring of $K[X_1,\ldots,X_n]/I$, so $K(X_1,\ldots,X_m)$ is a subfield of $L$.
Now comes the part I don't completely understand:
Choose $m<i\le n$ and examine $X_i$. Then there exists a nonzero $f\in K[X_1,\ldots,X_m][Y]\cap I$ such that $f(X_i) \in I$, where $f$ can be assumed irriducible because $I$ is prime. Therefore if $\alpha$ is a zero of $f$, then $\psi:K(X_1,\ldots, X_m,\alpha)\rightarrow L$, where $\psi(\alpha)=X_i$ and $\psi$ is the identity on the rest, is an embedding.
I don't understand why $\psi$ is necessarily an embedding, since if $x\in K(X_1,\ldots, X_m,\alpha)$ and $\psi(x) \in I$, then not necessarily $\psi(x) = f$, since $I\cap K[X_1,\ldots,X_m][Y]$ contains more polynomials than just $f$. Therefore $x$ is not necessarily $f(\alpha) = 0$, making $\psi$ not injective. What do I miss? Could it be true that $I = (f)$, which solves the problem?
Thank you for taking the effort if you've reached here!